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xxMikexx [17]
2 years ago
9

Fatima saved $15,000 toward a down payment on a house. She makes $5,375 a month. What is the maximum loan she can take on a hous

e?
A) $129,000
B) $193,500
C) $115,500
D) $144,000

Fatima has saved $15,000 toward a down payment on a house. She makes $5,375 a month. What is the maximum she can spend on a house?

A) $129,000
B) $193,500
C) $115,500
D) $144,000

Can anyone help me please.
Mathematics
2 answers:
Fittoniya [83]2 years ago
7 0

We will use \frac{28}{36} ratio to determine how much can Fatima afford as a loan for her house.

Let gross income of Fatima = $ x

Amount earned by Fatima monthly= $ 5,375

36 % of 5,375=

     \frac{36}{100}\times 5,375=1935$

Maximum amount that can be paid as a loan monthly = $ 1935

Total money that she can pay yearly for house loan if Fatima allowable money for paying loans is $ 1935=

       1935 \times 12=23220$

Money saved as a down payment = $ 15,000

So, maximum amount loan Fatima can take on a house= 23220 + 15000=$38,220

She is not allowed for loan for house if the cost of house is any of the four options.

(B) 28 % of $ 5375=

   \frac{28}{100}\times 5,375=1505$

Maximum amount that Fatima spend on her house monthly= $ 1505

Amount that fatima can spend yearly on her house = 1505× 12 =$18,060

For the four options given , none of the option matches Fatima's expectations.

Ivenika [448]2 years ago
5 0
It can be C or D because she can afford both. I'm guessing you go with C for now.



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Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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Part a

Description in words of the parameter p

p represent the real population proportion of people who showed partial or complete resistance to the antibiotic

\hat p represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

z_{\alpha/2} represent the critical value for the margin of error

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{973}{1714}=0.568 represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

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0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.  

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